A force is applied to a body of mass 2 ,kg at | vedclass

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(A) The area under the $F-t$ graph represents the impulse, which is equal to the change in momentum of the body.Since the body is initially at rest, t

Vedclass · Accepted Answer

$7.5$

Vedclass · Answer

(A) The area under the $F-t$ graph represents the impulse, which is equal to the change in momentum of the body.Since the body is initially at rest, the impulse equals the final momentum of the body after $1 \,s$.$	ext{Impulse} = 	ext{Area under } F-t 	ext{ graph}$$	ext{Impulse} = (10 \,N 	imes 0.5 \,s) + (20 \,N 	imes 0.5 \,s)$$	ext{Impulse} = 5 \,N-s + 10 \,N-s = 15 \,N-s$Since $	ext{Impulse} = \Delta p = m 	imes v - m 	imes u$ and $u = 0$, we have:$15 = 2 \,kg 	imes v$$v = \frac{15}{2} = 7.5 \,m/s$

$A$ force is applied to a body of mass $2 \,kg$ at rest on a frictionless horizontal surface, as shown in the force-time $(F-t)$ graph. The speed of the body after $1 \,s$ is: (in $\,m/s$)

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